Are resolvents for self-adjoint operators themselves self-adjoint?

Let [itex]T[/itex] be a (possibly unbounded) self-adjoint operator on a Hilbert space [itex]\mathscr H[/itex] with domain [itex]D(T)[/itex], and let [itex]\lambda \in \rho(T)[/itex]. Then we know that [itex](T-\lambda I)^{-1}[/itex] exists as a bounded operator from [itex]\mathscr H[/itex] to [itex]D(T)[/itex]. Question: do we also know that [itex](T-\lambda I)^{-1}[/itex] is self-adjoint? Can someone prove or give a counterexample?

Note: I have reason to believe this is true in the case that [itex]T[/itex] is a positive operator (i.e., [itex](Tx,x) \geq 0[/itex] for all [itex]x\in D(T)[/itex]). Whether it is true in general...well, I'm not sure!